Difference between revisions of "1992 AHSME Problems/Problem 25"
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\text{(E) } \frac{10}{\sqrt{3}}</math> | \text{(E) } \frac{10}{\sqrt{3}}</math> | ||
− | == Solution == | + | == Solution 1 (Extending Line Segments) == |
− | <math>\ | + | We begin by drawing a diagram. |
+ | <asy> | ||
+ | import olympiad; import cse5; import geometry; size(150); | ||
+ | defaultpen(fontsize(10pt)); | ||
+ | defaultpen(0.8); | ||
+ | dotfactor = 4; | ||
+ | pair A = origin; | ||
+ | pair C = A+dir(55); | ||
+ | pair D = A+dir(0); | ||
+ | pair B = extension(A,A+dir(90),C,C+dir(-155)); | ||
+ | label("$A$",A,S); | ||
+ | label("$C$",C,NE); | ||
+ | label("$D$",D,SE); | ||
+ | label("$B$",B,NW); | ||
+ | label("$4$",B--C,NW); | ||
+ | label("$3$",A--B,W); | ||
+ | draw(A--C--D--cycle); | ||
+ | draw(A--B--C); | ||
+ | draw(rightanglemark(B,C,D,2)); | ||
+ | draw(rightanglemark(B,A,D,2)); | ||
+ | </asy> | ||
+ | We extend <math>CB</math> and <math>DA</math> to meet at <math>E.</math> This gives us a couple right triangles in <math>CED</math> and <math>BEA.</math> | ||
+ | <asy> | ||
+ | import olympiad; import cse5; import geometry; size(250); | ||
+ | defaultpen(fontsize(10pt)); | ||
+ | defaultpen(0.8); | ||
+ | dotfactor = 4; | ||
+ | pair A = origin; | ||
+ | pair C = A+dir(55); | ||
+ | pair D = A+dir(0); | ||
+ | pair B = extension(A,A+dir(90),C,C+dir(-155)); | ||
+ | pair E = extension(A,A+2*dir(180),B,B+2*dir(-155)); | ||
+ | label("$A$",A,S); | ||
+ | label("$C$",C,NE); | ||
+ | label("$D$",D,SE); | ||
+ | label("$B$",B,NW); | ||
+ | label("$4$",B--C,NW); | ||
+ | label("$3$",A--B,W); | ||
+ | label("$E$",E,SW); | ||
+ | draw(A--C--D--cycle); | ||
+ | draw(A--B--C); | ||
+ | draw(rightanglemark(B,C,D,2)); | ||
+ | draw(rightanglemark(B,A,D,2)); | ||
+ | draw(A--E--B,dashed); | ||
+ | </asy> | ||
+ | We see that <math>\angle E = 30^\circ</math>. Hence, <math>\triangle BEA</math> and <math>\triangle DEC</math> are 30-60-90 triangles. | ||
+ | |||
+ | Using the side ratios of 30-60-90 triangles, we have <math>BE=2BA=6</math>. This tells us that <math>CE=BC+BE=4+6=10</math>. Also, <math>EA=3\sqrt{3}</math>. | ||
+ | |||
+ | Because <math>\triangle DEC\sim\triangle BEA</math>, we have <cmath>\frac{10}{3\sqrt{3}}=\frac{CD}{3}.</cmath> | ||
+ | Solving the equation, we have | ||
+ | <cmath>\begin{align*} | ||
+ | \frac{CD}3&=\frac{10}{3\sqrt{3}}\ | ||
+ | CD&=3\cdot\frac{10}{3\sqrt{3}}\ | ||
+ | CD&=\frac{10}{\sqrt{3}}\ | ||
+ | \end{align*}</cmath> | ||
+ | Hence, <math>CD=\boxed{\textbf{E}}</math>. | ||
+ | |||
+ | == Solution 2 (Cyclic Quadrilaterals, Right Triangles, LoC) == | ||
+ | |||
+ | Since <math>\angle{A}+\angle{C} = 180^{\circ}, ABCD</math> is cyclic. Using Ptolemy's Theorem gets <math>\text{(1): } 4AD + 3CD = \sqrt{37}BD.</math> | ||
+ | |||
+ | Right triangles <math>\Delta ABD</math> and <math>\Delta CBD</math> obtain <math>\text{(2): } 9+AD^2=BD^2</math> and <math>\text{(3):} 16+CD^2=BD^2,</math> respectively. | ||
+ | |||
+ | Seeing squares in <math>\text{(2)}</math> and <math>\text{(3)}</math>, we square <math>\text{(1)}</math> and get <math>\text{(4): } 16AD^2+9CD^2 + 24AD\cdot CD = 37BD^2.</math> | ||
+ | |||
+ | We don't like that <math>AD\cdot CD</math> term, but fortunately LoC exists: <math>37 = AD^2 + CD^2 - 2*AD*CD*\cos(60^{\circ})</math>. Solving for <math>AD\cdot CD</math> and plugging it into <math>\text{(4)}</math>, and using <math>AD^2 = 7 + CD^2</math> and <math>BD^2 = 16 + CD^2</math> from the first two equations, gets <math>40(CD^2+7)+33CD^2 - 24\cdot 37 = 37(16+CD^2).</math> | ||
+ | |||
+ | Solve for <math>CD = \boxed{\textbf{E}}</math>. | ||
+ | |||
+ | ~PureSwag | ||
== See also == | == See also == |
Latest revision as of 15:23, 24 October 2022
Contents
[hide]Problem
In , and . If perpendiculars constructed to at and to at meet at , then
Solution 1 (Extending Line Segments)
We begin by drawing a diagram. We extend and to meet at This gives us a couple right triangles in and We see that . Hence, and are 30-60-90 triangles.
Using the side ratios of 30-60-90 triangles, we have . This tells us that . Also, .
Because , we have Solving the equation, we have Hence, .
Solution 2 (Cyclic Quadrilaterals, Right Triangles, LoC)
Since is cyclic. Using Ptolemy's Theorem gets
Right triangles and obtain and respectively.
Seeing squares in and , we square and get
We don't like that term, but fortunately LoC exists: . Solving for and plugging it into , and using and from the first two equations, gets
Solve for .
~PureSwag
See also
1992 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Problem 26 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
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